Answer to Question #155724 in General Chemistry for Willy

Question

In the production of Zinc sulfide, 36.8 g of zinc is made to react with 19.4 g of sulfur. 

𝑍𝑛 + 𝑆 → 𝑍𝑛𝑆

a. How many moles of ZnS is produced when sulfur is completely used up?

b. How many grams of ZnS is produced when zinc is completely used up?

c. Which reactant is the limiting reagent?

d. How many grams of the excess reagent is left?

Answer 1

(a) Moles of Zinc sulfide produced by 19.4g of Sulphur.

First, find moles of Sulfur used;

Moles sulphur $=\dfrac{mass of sulphur}{molar massof sulphur} = \dfrac{19.4g}{32g/m ol} = 0.606 mol$

Moles ratio of Sulphur:ZnS = 1:1

Thus moles of ZnS $=\dfrac{1}{1}x0.606 = 0.606mol$

(b) Grams of ZnS produced by 36.8g of Zn

Moles of Zn used $=\dfrac{mass of Zn}{molar massof Zn} = \dfrac{36.8g}{65.38g/mol} = 0.5629mol$

Moles of ZnS produced, thus

The mole ratio of Zn: ZnS = 1:1

Moles of ZnS $=\dfrac{1}{1}x0.5629mol = 0.5629 mol$

Mass of ZnS = Moles of ZnS x RFM of ZnS

= 0.5629 mol x 97.474g/mol

= 54.86g

(c) Limiting reagent

Moles of sulphur = 0.606mol

Moles of zinc, therefore, required in the 1:1 ratio =0.606mol

But there are only 0.5629mol of Zn;

Therefore, Zinc is the limiting reagent

(d) Grams of excess reagent (sulfur)

Moles of Zn = 0.5629 mol

Moles of sulfur that react is, therefore, 0.5629mol according to the 1:1 ratio

Excess sulfur = 0.606mol - 0.5629

= 0.037mol